Joint detection probabilities in entangled photon pairs
In my previous post (here), I introduced entangled photon pairs and demonstrated their non-factorizable state. Now, let’s explore the joint probabilities of measuring different polarization outcomes for these entangled photons.
Consider the entangled state of the photon pair:
| \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle = \frac{1}{\sqrt{2}} \left(| \mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle\right)
We want to calculate the probability of obtaining specific outcomes when measuring the polarization of each photon. Let’s denote the probability of measuring +1 polarization for photon \nu_1 and +1 for photon \nu_2 (with polarizers oriented at angles \alpha and \beta respectively) as \mathcal P_{++}(\mathbf a, \mathbf b). This is calculated by projecting the entangled state onto the joint measurement state |+_{\mathbf a}, +_{\mathbf b}\rangle:
\mathcal P_{++}(\mathbf a, \mathbf b) = |\langle +_{\mathbf a}, +_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle|^2
After expanding | +_{\mathbf a}, +_{\mathbf b} \rangle = (\cos(\alpha) | \mathbf x_1 \rangle + \sin(\alpha) | \mathbf y_1 \rangle) \otimes (\cos(\beta) | \mathbf x_2 \rangle + \sin(\beta) | \mathbf y_2 \rangle) and projecting, we find:
\mathcal P_{++}(\mathbf a, \mathbf b) = \frac{1}{2} \cos^2(\alpha - \beta)
Similarly, we can calculate the probability \mathcal P_{--}(\mathbf a, \mathbf b) for obtaining -1 for both photons:
\mathcal P_{--}(\mathbf a, \mathbf b) = |\langle -_{\mathbf a}, -_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle|^2 = \frac{1}{2} \cos^2(\alpha - \beta)
For the probabilities of mixed outcomes, \mathcal P_{+-}(\mathbf a, \mathbf b) ( +1 for photon \nu_1, -1 for photon \nu_2) and \mathcal P_{-+}(\mathbf a, \mathbf b) ( -1 for photon \nu_1, +1 for photon \nu_2), we obtain:
\mathcal P_{+-}(\mathbf a, \mathbf b) = |\langle +_{\mathbf a}, -_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle|^2 = \frac{1}{2} \sin^2(\alpha - \beta)
\mathcal P_{-+}(\mathbf a, \mathbf b) = |\langle -_{\mathbf a}, +_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle|^2 = \frac{1}{2} \sin^2(\alpha - \beta)
Notice that \mathcal P_{++}(\mathbf a, \mathbf b) = \mathcal P_{--}(\mathbf a, \mathbf b) and \mathcal P_{+-}(\mathbf a, \mathbf b) = \mathcal P_{-+}(\mathbf a, \mathbf b). Furthermore, the probabilities depend on the difference between the angles of the polarizers, \alpha - \beta, highlighting the correlation between the two photons’ polarizations. These joint probabilities demonstrate the strong correlations inherent in entangled states, a key feature that distinguishes quantum mechanics from classical physics.
For more insights into this topic, you can find the details here.